You should be able to figure out that the graph of y=1/x^2 looks like a tall tower at x=0, and tails off to y=0 as x gets farther from 0. There is a vertical asymptote at x=0, and a horizontal asumptote at y=0. The domain is all reals except x=0, and the range is all reals > 0.
Now, if we replace x by x+4, then all the values on the graph occur at a value of x which is 4 less than originally. That is, the graph is shifted left by 4 units. Now the v.a. is at x = -4. The domain is now all reals except x = -4. The new graph is y=1/(x+4)^2
Now flip it upside down by replacing y with -y. The range is now all reals <0. The graph is now y = -1/(x+4)^2
Finally, shift the graph upward by adding 6 to each y. Now the range is all reals < 6. Now the graph is y=6-1/(x+4)^2
Visit wolframalpha.com and type in 1/x^2 to see the original function, and then apply these transformations one by one to see how the graph changes.
Graph the function f(x)=6-1/(x+4)^2 using transformations.
State:
a.the domain
b.the range
c.the asymptotes
i do not get this at all, i am totally lost! please help!!!
1 answer
1 answer